Two-sided competition, platform services and online shopping market structure

Abstract

On an online shopping platform, for users on one side (buyers or sellers), the affiliation decisions depend not only on their stand-alone valuations for the platform’s services, but also on the platform’s user base on the other side. This paper investigates online shopping market structure based on platform services on the two sides and cross-group network effects. We find that the entry of inactive platforms hinges on the side where the strength gap between active platforms and inactive platforms is smaller, i.e. the side on which the average stand-alone valuation of users for platform services is relatively high. Specifically, given the sum of the average stand-alone valuations on the two sides, the market structure is more competitive when the proportion of the average stand-alone valuation on one side to the other is too high or too low. Furthermore, the market structure is more monopolistic when cross-group network effects are strong. We also find that social welfare and the total surplus of users on the two sides may both be lower when the market structure is more competitive. In a case where users on one side multi-home, we find that inactive platforms should pay attention to the average stand-alone valuation on the multi-homing side to enter the markets, whereas active platforms should pay attention to the average stand-alone valuation on the single-homing side to prevent the entry of inactive platforms.

Appendices

Appendix 1

We build a dynamic model following Evans and Schmalensee (2010) to analyze the stability condition as follows:

$$\begin{array}{*{20}l} {\dot{x}^{i} (t) = \frac{{dx^{i} (t)}}{dt} = h^{x} (d^{ib} (t) - x^{i} (t))} \hfill \\ {\dot{y}^{i} (t) = \frac{{dy^{i} (t)}}{dt} = h^{y} (d^{is} (t) - y^{i} (t))} \hfill \\ \end{array} ,$$

where $x^{i} (t)$ and $y^{i} (t)$, respectively, denote the numbers of buyers and sellers on platform $i$ at time $t$, and $h^{x}$ and $h^{y}$ are increasing functions with $h^{x} (0) = 0$ and $h^{y} (0) = 0$. In addition, given $x^{i} (t)$ and $y^{i} (t)$, from Eqs. (3–6), the demand of buyers denoted by $d^{ib} (t)$ and the demand of sellers denoted by $d^{is} (t)$ on platform $i$ at time $t$ satisfy

$$d^{ib} (t) = (\overline{B} + v(y^{i} (t)) - p^{ib*} )/\sigma^{b} - Z^{b* - }$$

and

$$d^{is} (t) = (\overline{S} + u(x^{i} (t)) - p^{is*} )/\sigma^{s} - Z^{s* - } ,$$

respectively.

The linear approximation of the dynamic model then is

$$\frac{d}{dt}\left[ {\begin{array}{ll} {x^{i} } \\ {y^{i} } \\ \end{array} } \right] = \left[ {\begin{array}{ll} a & b \\ c & d \\ \end{array} } \right]\left[ {\begin{array}{ll} {x^{i} } \\ {y^{i} } \\ \end{array} } \right],$$

where $a = - \left. {(h^{x} )^{\prime}} \right|_{{(x^{i} ,y^{i} ) = (n^{ib*} ,n^{is*} )}}$, $b = (h^{x} )^{\prime}d_{{y^{i} }}^{ib} |_{{(x^{i} ,y^{i} ) = (n^{ib*} ,n^{is*} )}}$, $c = (h^{y} )^{\prime}d_{{x^{i} }}^{is} |_{{(x^{i} ,y^{i} ) = (n^{ib*} ,n^{is*} )}}$ and $d = - (h^{y} )^{\prime}|_{{(x^{i} ,y^{i} ) = (n^{ib*} ,n^{is*} )}}$.

According to the stability theory of differential equations, the stable symmetric equilibrium should satisfy

$$p \triangleq - (a + d) = (h^{x} )^{\prime} + (h^{y} )^{\prime} > 0$$

and

$$q \triangleq ad - bc = (h^{x} )^{\prime}(h^{y} )^{\prime}(1 - \frac{{v^{\prime}(n^{is*} )u^{\prime}(n^{ib*} )}}{{\sigma^{b} \sigma^{s} }}) > 0.$$

Since $h^{x}$ and $h^{y}$ are increasing functions, we have $p > 0$. The stability condition then only requires $u^{\prime}(n^{ib*} )v^{\prime}(n^{is*} ) < \sigma^{b} \sigma^{s}$.

Appendix 2

Proof of Lemma 1

From $\pi_{{n^{ib} }}^{i} = \pi_{{n^{is} }}^{i} = 0$ and the rationality condition, we have

$$\begin{array}{*{20}l} {\pi_{{n^{ib} }}^{i} = \overline{B} + v(n^{is*} ) - \sigma^{b} Z^{b*} - \sigma^{b} n^{ib*} = 0} \hfill \\ {\pi_{{n^{is} }}^{i} = \overline{S} + u(n^{ib*} ) - \sigma^{s} Z^{s*} - \sigma^{s} n^{is*} = 0} \hfill \\ \end{array} .$$

Taking the sum of the above equations over $i \in \Omega$, we obtain

$$\sum\limits_{i \in \Omega } {\pi_{{n^{ib} }}^{i} } = k\overline{B} + \sum\limits_{i \in \Omega } {v(n^{{is{*}}} )} - k\sigma^{b} Z^{b*} - \sigma^{b} Z^{b*} = k\overline{B} + \sum\limits_{i \in \Omega } {v(n^{{is{*}}} } ) - \sigma^{b} (k + 1)\left(\frac{{\overline{B} + v(n^{{is{*}}} ) - p^{ib*} }}{{\sigma^{b} }}\right) = 0$$

and

$$\sum\limits_{i \in \Omega } {\pi_{{n^{is} }}^{i} = } k\overline{S} + \sum\limits_{i \in \Omega } {u(n^{{ib{*}}} )} - k\sigma^{s} Z^{s*} - \sigma^{s} Z^{s*} = k\overline{S} + \sum\limits_{i \in \Omega } {u(n^{{ib{*}}} )} - \sigma^{s} (k + 1)\left(\frac{{\overline{S} + u(n^{{ib{*}}} ) - p^{is*} }}{{\sigma^{s} }}\right) = 0.$$

Simplifying the two equations above, we have Lemma 1. $\square$

Proof of Lemma 2

From Eqs. (5–6) and (10–11), in the symmetric equilibria, we can obtain $p^{bk*} = \sigma^{b} n^{bk*}$ and $p^{sk*} = \sigma^{s} n^{sk*}$. Hence, $\pi^{k*} = \sigma^{b} (n^{bk*} )^{2} + \sigma^{s} (n^{sk*} )^{2}$ and we only need to verify the signs of $n_{\zeta }^{bk*}$ and $n_{\zeta }^{sk*}$, where $\zeta = B$, $S$ or $k$.

• (1) $\zeta = B$ or $S$

According to the concavity of $u( \cdot )$ and $v( \cdot )$, in the symmetric equilibrium, we obtain that the partial derivative of $n^{bk*}$ to $n^{sk*}$ in Eq. (10) is lower than that in Eq. (11) as follows:

$$\frac{{v^{\prime}(n^{sk*} )}}{{\sigma^{b} (k + 1)}} < \frac{{\sigma^{s} (k + 1)}}{{u^{\prime}(n^{bk*} )}}.$$

(15)

Taking the first derivative of Eqs. (10–11) with respect to $B$ and $S$, we obtain

$$\begin{array}{*{20}l} {n_{B}^{bk*} = \frac{{(k + 1)\sigma^{s} }}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}} > 0} \hfill \\ {n_{S}^{bk*} = \frac{{v^{\prime}(n^{sk*} )}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}} > 0} \hfill \\ \end{array} ,$$

(16)

since $\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} ) > 0$ according to Eq. (15). Similarly, due to the symmetry of two sides, we can prove that $n_{B}^{sk*} > 0$ and $n_{S}^{sk*} > 0$.

• (2) $\zeta = k$

Taking the first derivative of Eqs. (10–11) with respect to $k$, we obtain

$$\left[ {\begin{array}{ll} { - \sigma^{b} (k + 1)} & {v^{\prime}(n^{sk*} )} \\ {u^{\prime}(n^{bk*} )} & { - \sigma^{s} (k + 1)} \\ \end{array} } \right]\left[ {\begin{array}{ll} {n_{k}^{bk*} } \\ {n_{k}^{sk*} } \\ \end{array} } \right] = \left[ {\begin{array}{ll} {\sigma^{b} n^{bk*} } \\ {\sigma^{s} n^{sk*} } \\ \end{array} } \right].$$

(17)

Simplifying the equations above, we obtain

$$n_{k}^{bk*} = \frac{{ - \sigma^{b} \sigma^{s} (k + 1)n^{bk*} - \sigma^{s} n^{sk*} v^{\prime}(n^{sk*} )}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}} < 0$$

and

$$n_{k}^{sk*} = \frac{{ - \sigma^{s} \sigma^{b} (k + 1)n^{sk*} - \sigma^{b} n^{bk*} u^{\prime}(n^{bk*} )}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}} < 0.$$

$\square$

Proof of Lemma 3

Taking the first derivative of Eqs. (10–11) with respect to $\sigma^{b}$, we obtain

$$\left[ {\begin{array}{ll} {\sigma^{b} (k + 1)} & { - v^{\prime}(n^{sk*} )} \\ {u^{\prime}(n^{bk*} )} & { - \sigma^{s} (k + 1)} \\ \end{array} } \right]\left[ {\begin{array}{ll} {n_{{\sigma^{b} }}^{bk*} } \\ {n_{{\sigma^{b} }}^{sk*} } \\ \end{array} } \right] = \left[ {\begin{array}{ll} {0.5 - (k + 1)n^{bk*} } \\ 0 \\ \end{array} } \right]$$

Then we obtain

$$n_{{\sigma^{b} }}^{bk*} = \frac{{\sigma^{s} [0.5 - (k + 1)n^{bk*} ](k + 1)}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}}$$

and

$$n_{{\sigma^{b} }}^{sk*} = \frac{{[0.5 - (k + 1)n^{bk*} ]u^{\prime}(n^{bk*} )}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}}.$$

Since $\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} ) > 0$ according to Eq. (15), we obtain that ${\text{sgn}} (n_{{\sigma^{b} }}^{bk*} ) = {\text{sgn}} (n_{{\sigma^{b} }}^{sk*} ) = {\text{sgn}} (0.5 - Z^{b*} - n^{bk*} )$, and $n^{bk*}$ and $n^{sk*}$ are monotonic functions on $\sigma^{b}$.

Next, since $p^{sk*} = \sigma^{s} n^{sk*}$, we have ${\text{sgn}} (p_{{\sigma^{b} }}^{sk*} ) = {\text{sgn}} (n_{{\sigma^{b} }}^{sk*} )$. For $p^{bk*} = \sigma^{b} n^{bk*}$, if $Z^{b*} + n^{bk*} < 0.5$, we have $p_{{\sigma^{b} }}^{bk*} > 0$ and if $Z^{b*} + n^{bk*} \ge 0.5$, by Eq. (9), we obtain

$$\begin{aligned} p_{{\sigma^{b} }}^{bk*} & = n^{bk*} + \frac{{\sigma^{b} \sigma^{s} [0.5 - (k + 1)n^{bk*} ](k + 1)}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}} \\ & > n^{bk*} + \frac{{\sigma^{b} \sigma^{s} [0.5 - (k + 1)n^{bk*} ](k + 1)}}{{\sigma^{b} \sigma^{s} (k^{2} + 2k)}}. \\ \end{aligned}$$

Simplify the equation above, we have $p_{{\sigma^{b} }}^{bk*} > \frac{{0.5(k + 1) - n^{bk*} }}{{k^{2} + 2k}} > 0$, and then we can prove that $p_{{\sigma^{b} }}^{bk*}$ is always positive.

Finally, for $\pi^{k*} = \sigma^{b} (n^{bk*} )^{2} + \sigma^{s} (n^{sk*} )^{2}$, when $Z^{b*} + n^{bk*} \le 0.5$, we obtain $\pi_{{\sigma^{b} }}^{k*} > 0$ as $n_{{\sigma^{b} }}^{bk*} \ge 0$ and $n_{{\sigma^{b} }}^{sk*} \ge 0$. However, when $Z^{l*} + n^{lk*} > 0.5$, $\pi^{k*}$ may decrease with $\sigma^{b}$ as $n_{{\sigma^{b} }}^{bk*} < 0$ and $n_{{\sigma^{b} }}^{sk*} < 0$ (see Footnote 9).

Similarly, due to the symmetry of two sides, we can prove others. $\square$

Proof of Proposition 1

From Eqs. (10–11), since $Z^{b*} { = }kn^{bk*}$ and $Z^{s*} { = }kn^{sk*}$, we have

$$\begin{array}{*{20}l} {\sigma^{b} (k + 1)\frac{{Z^{b*} }}{k} = \overline{B} + v\left( {\frac{{Z^{s*} }}{k}} \right)} \hfill \\ {\sigma^{s} (k + 1)\frac{{Z^{s*} }}{k} = \overline{S} + u\left( {\frac{{Z^{b*} }}{k}} \right)} \hfill \\ \end{array} .$$

Taking the first derivative of the above equations with respect to $k$, we have

$$\begin{array}{*{20}l} {\frac{{\sigma^{b} (k + 1)}}{k}Z_{k}^{b*} - \frac{{\sigma^{b} }}{{k^{2} }}Z^{b*} = v^{\prime}\left( {\frac{{Z^{s*} }}{k}} \right)\frac{{kZ_{k}^{s*} - Z^{s*} }}{{k^{2} }}} \hfill \\ {\frac{{\sigma^{s} (k + 1)}}{k}Z_{k}^{s*} - \frac{{\sigma^{s} }}{{k^{2} }}Z^{s*} = u^{\prime}\left( {\frac{{Z^{b*} }}{k}} \right)\frac{{kZ_{k}^{b*} - Z^{b*} }}{{k^{2} }}} \hfill \\ \end{array} .$$

Solving the equations above, we obtain

$$Z_{k}^{b*} = \frac{{k(\sigma^{s} \sigma^{b} n^{bk*} - \sigma^{s} n^{sk*} v^{\prime}(n^{sk*} )) + n^{bk*} \Delta }}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}}$$

and

$$Z_{k}^{s*} = \frac{{k\sigma^{b} (\sigma^{s} n^{sk*} - n^{bk*} u^{\prime}(n^{bk*} )) + n^{sk*} \Delta }}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}},$$

where $\Delta = \sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} ) > 0$ by Eq. (9).

Since ${\text{sgn}} (Z_{k}^{l*} ) = {\text{sgn}} (CS_{k}^{l*} )$, we only need to verify the signs of $Z_{k}^{l*}$, $l = b,s$. Note that the sign of $Z_{k}^{l*}$, $l = b,s$ is uncertain (see Fig. 1a). However, if $Z_{k}^{b*} < 0$, we can obtain $Z_{k}^{s*} > 0$ since $\sigma^{s} n^{sk*} v^{\prime}(n^{sk*} ) > \sigma^{s} \sigma^{b} n^{bk*} > u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} )n^{bk*}$ (by $\Delta > 0$). We then obtain the first results. The second results can be proved by Fig. 1b directly.

Next, we discuss the change of $CS^{b}$ and $CS^{s}$ with respect to $k$. Firstly, note that if $Z_{k}^{b*} < 0$, the numerator of $Z_{k}^{b*}$, i.e. $\sigma^{s} \sigma^{b} Z^{b*} - \sigma^{s} Z^{s*} v^{\prime}(n^{sk*} ) + n^{bk*} \Delta$, decreases with $k$ as $Z_{k}^{s*} > 0$, $n_{k}^{bk*} < 0$ and $n_{k}^{sk*} < 0$. Hence, when $\left. {Z_{k}^{b*} } \right|_{{k = \tilde{k}}} < 0$ for some $\tilde{k} \le \overline{k}$, we have $Z_{k}^{b*} < 0$ and $Z_{k}^{s*} > 0$ for $k \in [\tilde{k},\overline{k}]$, where $\Delta = 0$ if $k = \overline{k}$.¹⁷ Secondly, we can prove that ${\text{sgn}} (\left. {Z_{k}^{b*} } \right|_{{k = \overline{k}}} ) = {\text{sgn}} ( - \left. {Z_{k}^{s*} } \right|_{{k = \overline{k}}} )$ since

$$\left. {Z_{k}^{b*} } \right|_{{k = \overline{k}}} = \frac{{k\sigma^{s} (\sigma^{b} n^{bk*} - n^{sk*} v^{\prime}(n^{sk*} ))}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}}$$

and

$$\left. {Z_{k}^{s*} } \right|_{{k = \overline{k}}} = \frac{{ku^{\prime}(n^{bk*} )((v^{\prime}(n^{sk*} )n^{sk*} - \sigma^{b} n^{bk*} )}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}}.$$

Therefore, for the change of $CS^{b}$ and $CS^{s}$ with respect to $k$, three cases might exist. Case 1: The surplus of users on one side first increases with $k$ and then decreases with $k$, and the surplus of users on the other side increases with $k$. One example of Case 1 is shown in Fig. 1. Case 2: The surplus of users on one side decreases with $k$, and the surplus of users on the other side increases with $k$. One example of Case 2 is that $\sigma^{b} = \sigma^{s} = 4$, $B = S = - 2$, $\gamma^{b} = 0.9$, $\gamma^{s} = 0.1$ and $\lambda^{b} = \lambda^{s} = 1$. Case 3: The surplus of users on either side increases with $k$. One example of Case 3 is shown in Fig. 2. In this example, we also have $\left. {Z_{k}^{b*} } \right|_{{k = \overline{k}}} = \left. {Z_{k}^{s*} } \right|_{{k = \overline{k}}} = 0$. $\square$

Proof of Corollary 1

According to the proof of Proposition 1, when $B = S$, $\sigma^{b} = \sigma^{s}$ and $u( \cdot ) = v( \cdot )$, we obtain

$$Z_{k}^{b*} = \frac{{\sigma^{b} kn^{bk*} (\sigma^{b} - u^{\prime}(n^{bk*} )) + n^{bk*} \Delta }}{{(\sigma^{b} )^{2} (k + 1)^{2} - (u^{\prime}(n^{bk*} ))^{2} }}.$$

Since $\Delta = \sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} ) > 0$, we have $\sigma^{b} - u^{\prime}(n^{bk*} ) > 0$ and then $Z_{k}^{b*} > 0$. The result for social welfare can be proved by Fig. 2b directly. $\square$

Proof of Proposition 2

For $(S,B)$ on curve $\overline{B} = \sigma^{b} Z^{b*}$, from Eqs. (10–11), we obtain

$$\begin{array}{*{20}l} {\sigma^{b} (k + 1)\frac{{\overline{B}}}{{k\sigma^{b} }} = \overline{B} + v(n^{sk*} )} \hfill \\ {\sigma^{s} (k + 1)n^{sk*} = \overline{S} + u\left( {\frac{{\overline{B}}}{{k\sigma^{b} }}} \right)} \hfill \\ \end{array} .$$

Simplifying the equations above, we have

$$\overline{S} = \sigma^{s} (k + 1)v^{ - 1} \left( {\frac{{\overline{B}}}{k}} \right) - u\left( {\frac{{\overline{B}}}{{\sigma^{b} k}}} \right).$$

Since $Z_{S}^{b*} > 0$, we obtain that $\overline{B} - \sigma^{b} Z^{b*} \le 0$ is equivalent to

$$\overline{S} \ge \sigma^{s} (k + 1)v^{ - 1} \left( {\frac{{\overline{B}}}{k}} \right) - u\left( {\frac{{\overline{B}}}{{\sigma^{b} k}}} \right).$$

Similarly, due to the symmetry of two sides, we can prove that $\overline{S} - \sigma^{s} Z^{s*} \le 0$ is equivalent to $\overline{B} \ge \sigma^{b} (k + 1)u^{ - 1} (\frac{{\overline{S}}}{k}) - v(\frac{{\overline{S}}}{{\sigma^{s} k}})$. $\square$

Proof of Proposition 3

According to the proof of Proposition 1, we have

$$Z_{k}^{b*} = \frac{{k\sigma^{s} (\sigma^{b} n^{bk*} - n^{sk*} v^{\prime}(n^{sk*} )) + n^{bk*} \Delta }}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - v^{\prime}(n^{sk*} )u^{\prime}(n^{bk*} )}},$$

where $\Delta = \sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} ) > 0$.

For the points on curve $B = f(S,k)$, we can obtain $\sigma^{b} n^{bk*} = v(n^{sk*} )$ by Eq. (10) as $\overline{B} = \sigma^{b} Z^{b*} = k\sigma^{b} n^{bk*}$. Then we have

$$\sigma^{b} n^{bk*} - n^{sk*} v^{\prime}(n^{sk*} ) = v(n^{sk*} ) - n^{sk*} v^{\prime}(n^{sk*} ) > v(0) = 0$$

since $v( \cdot )$ is a concave function.

With the analysis above, we finally obtain

$$Z_{k}^{b*} > \frac{{k\sigma^{s} (\sigma^{b} n^{bk*} - n^{sk*} v^{\prime}(n^{sk*} ))}}{{\sigma^{b} \sigma^{s} (k + 1)^{2} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} )}} > 0.$$

Hence, curve $B = f(S,k + 1)$ is on the left of curve $B = f(S,k)$. Similarly, we can prove that curve $S = g(B,k + 1)$ is below curve $S = g(B,k)$ (see Fig. 4). $\square$

Proof of Proposition 6

From Eq. (14), since $Z^{b*} { = }kn^{bk*}$ and $Z^{s*} { = }n^{sk*}$, we have

$$\begin{array}{*{20}l} {\sigma^{b} (k + 1)\frac{{Z^{b*} }}{k} = \overline{B} + v(Z^{s*} )} \hfill \\ {2\sigma^{s} Z^{s*} = \overline{S} + u\left( {\frac{{Z^{b*} }}{k}} \right)} \hfill \\ \end{array} .$$

Taking the first derivative of the above equations with respect to $k$, we have

$$\begin{array}{*{20}l} {\frac{{\sigma^{b} (k + 1)}}{k}Z_{k}^{b*} - \frac{{\sigma^{b} }}{{k^{2} }}Z^{b*} = v^{\prime}(Z^{s*} )Z_{k}^{s*} } \hfill \\ {2\sigma^{s} Z_{k}^{s*} = u^{\prime}\left( {\frac{{Z^{b*} }}{k}} \right)\frac{{kZ_{k}^{b*} - Z^{b*} }}{{k^{2} }}} \hfill \\ \end{array} .$$

Solving the equations above, we obtain

$$Z_{k}^{b*} = \frac{{n^{bk*} (2\sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} ))}}{{2(k + 1)\sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} )}} > 0$$

and

$$Z_{k}^{s*} = \frac{{ - \sigma^{b} n^{bk*} u^{\prime}(n^{bk*} )}}{{2(k + 1)\sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} )}} < 0,$$

since $\Delta = \sigma^{b} \sigma^{s} - u^{\prime}(n^{bk*} )v^{\prime}(n^{sk*} ) > 0$.

We then can prove that $CS_{k}^{b} > 0$ since $CS^{b} = \frac{1}{2}\sigma^{b} (Z^{b*} )^{2}$. However, the sign of $CS_{k}^{s} = \frac{1}{2}\sigma^{s} Z^{s*} (Z^{s*} + 2kZ_{k}^{s*} )$ is uncertain although the surplus of sellers from one active platform, i.e. $\frac{1}{2}\sigma^{s} (Z^{s*} )^{2}$, decreases with $k$ (see Fig. 6a). In addition, the results for the total surplus and social welfare can be proved by Fig. 6b directly. $\square$

Proof of Proposition 7

For $(S,B)$ on curve $\overline{B} = \sigma^{b} Z^{b*}$, from Eq. (14), we obtain

$$\begin{array}{*{20}l} {(k + 1)\sigma^{b} \frac{{\overline{B}}}{{k\sigma^{b} }} = \overline{B} + v(n^{sk*} )} \hfill \\ {2\sigma^{s} n^{sk*} = \overline{S} + u\left( {\frac{{\overline{B}}}{{k\sigma^{b} }}} \right)} \hfill \\ \end{array} .$$

Simplifying the equations above, we have

$$\overline{S} = 2\sigma^{s} v^{ - 1} \left( {\frac{{\overline{B}}}{k}} \right) - u\left( {\frac{{\overline{B}}}{{\sigma^{b} k}}} \right).$$

Since $Z_{S}^{b*} > 0$, we obtain that $\overline{B} - \sigma^{b} Z^{b*} \le 0$ is equivalent to

$$\overline{S} \ge 2\sigma^{s} v^{ - 1} \left( {\frac{{\overline{B}}}{k}} \right) - u\left( {\frac{{\overline{B}}}{{\sigma^{b} k}}} \right).$$

Similarly, due to the symmetry of two sides, we can prove that $\overline{S} - \sigma^{s} Z^{s*} (\Lambda ) \le 0$ is equivalent to $\overline{B} \ge \sigma^{b} (k + 1)u^{ - 1} (\overline{S}) - v\left( {\frac{{\overline{S}}}{{\sigma^{s} }}} \right)$. $\square$

Proof of Proposition 8

By comparing Proposition 2 with Proposition 7, we obtain

$$\sigma^{s} (k + 1)v^{ - 1} \left( {\frac{{\overline{B}}}{k}} \right) - u\left( {\frac{{\overline{B}}}{{\sigma^{b} k}}} \right) > 2\sigma^{s} v^{ - 1} \left( {\frac{{\overline{B}}}{k}} \right) - u\left( {\frac{{\overline{B}}}{{\sigma^{b} k}}} \right)$$

and

$$\sigma^{b} (k + 1)u^{ - 1} \left( {\frac{{\overline{S}}}{k}} \right) - v\left( {\frac{{\overline{S}}}{{\sigma^{s} k}}} \right) < \sigma^{b} (k + 1)u^{ - 1} (\overline{S}) - v\left( {\frac{{\overline{S}}}{{\sigma^{s} }}} \right)$$

for $k > 1$, since $B$ increases with $S$ on curve $S = g^{m} (B,k)$, i.e. $\sigma^{b} (k + 1)u^{ - 1} (\overline{S}) - v\left( {\frac{{\overline{S}}}{{\sigma^{s} }}} \right)$ increases with $S$.

Therefore, curve $B = f^{m} (S,k)$ is on the left of curve $B = f(S,k)$ and curve $S = g^{m} (B,k)$ is above curve $S = g(B,k)$. $\square$